Rank the increase in length from greatest to smallest.

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Answer 1

Question

Chapter 13, Problem 6P

To determine

Rank the increase in length from greatest to smallest.

Expert Solution & Answer

Answer to Problem 6P

The rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Explanation of Solution

Write an expression for the increase in length.

ΔL=LαΔT=Lα(Tf−Ti) (I)

Here, ΔL is the increase in length, L is the initial length, α is the coefficient of thermal expansion, ΔT is the change in temperature, Tf is final temperature and Ti is the initial temperature.

Conclusion:

Consider slab (a).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(20 K)=1.44×10−4 m

Consider slab (b).

Substitute 90 cm for L, 50°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(30 K)=2.16×10−4 m

Consider slab (c).

Substitute 60 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(8×10−6 K−1)(20 K)=0.96×10−4 m

Consider slab (d).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(12×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(12×10−6 K−1)(20 K)=2.16×10−4 m

Consider slab (e).

Substitute 60 cm for L, 50°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(12×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(12×10−6 K−1)(30 K)=2.16×10−4 m

Thus, the rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

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The force of the quadriceps (Fq) and force of the patellar tendon (Fp) is identical (i.e., 1000 N each). In the figure below angle in blue is Θ and the in green is half Θ (i.e., Θ/2). A) Calculate the patellar reaction force (i.e., R resultant vector is the sum of the horizontal component of the quadriceps and patellar tendon force) at the following joint angles: you need to provide a diagram showing the vector and its components for each part. a1) Θ = 160 degrees, a2) Θ = 90 degrees. NOTE: USE ONLY TRIGNOMETRIC FUNCTIONS (SIN/TAN/COS, NO LAW OF COSINES, NO COMPLICATED ALGEBRAIC EQUATIONS OR ANYTHING ELSE, ETC. Question A has 2 parts!

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Answer 2

Question

Chapter 13, Problem 6P

To determine

Rank the increase in length from greatest to smallest.

Expert Solution & Answer

Answer to Problem 6P

The rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Explanation of Solution

Write an expression for the increase in length.

ΔL=LαΔT=Lα(Tf−Ti) (I)

Here, ΔL is the increase in length, L is the initial length, α is the coefficient of thermal expansion, ΔT is the change in temperature, Tf is final temperature and Ti is the initial temperature.

Conclusion:

Consider slab (a).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(20 K)=1.44×10−4 m

Consider slab (b).

Substitute 90 cm for L, 50°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(30 K)=2.16×10−4 m

Consider slab (c).

Substitute 60 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(8×10−6 K−1)(20 K)=0.96×10−4 m

Consider slab (d).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(12×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(12×10−6 K−1)(20 K)=2.16×10−4 m

Consider slab (e).

Substitute 60 cm for L, 50°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(12×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(12×10−6 K−1)(30 K)=2.16×10−4 m

Thus, the rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

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Answer 3

Question

Chapter 13, Problem 6P

To determine

Rank the increase in length from greatest to smallest.

Expert Solution & Answer

Answer to Problem 6P

The rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Explanation of Solution

Write an expression for the increase in length.

ΔL=LαΔT=Lα(Tf−Ti) (I)

Here, ΔL is the increase in length, L is the initial length, α is the coefficient of thermal expansion, ΔT is the change in temperature, Tf is final temperature and Ti is the initial temperature.

Conclusion:

Consider slab (a).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(20 K)=1.44×10−4 m

Consider slab (b).

Substitute 90 cm for L, 50°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(30 K)=2.16×10−4 m

Consider slab (c).

Substitute 60 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(8×10−6 K−1)(20 K)=0.96×10−4 m

Consider slab (d).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(12×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(12×10−6 K−1)(20 K)=2.16×10−4 m

Consider slab (e).

Substitute 60 cm for L, 50°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(12×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(12×10−6 K−1)(30 K)=2.16×10−4 m

Thus, the rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

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Answer 4

Question

Chapter 13, Problem 6P

To determine

Rank the increase in length from greatest to smallest.

Expert Solution & Answer

Answer to Problem 6P

The rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Explanation of Solution

Write an expression for the increase in length.

ΔL=LαΔT=Lα(Tf−Ti) (I)

Here, ΔL is the increase in length, L is the initial length, α is the coefficient of thermal expansion, ΔT is the change in temperature, Tf is final temperature and Ti is the initial temperature.

Conclusion:

Consider slab (a).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(20 K)=1.44×10−4 m

Consider slab (b).

Substitute 90 cm for L, 50°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(30 K)=2.16×10−4 m

Consider slab (c).

Substitute 60 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(8×10−6 K−1)(20 K)=0.96×10−4 m

Consider slab (d).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(12×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(12×10−6 K−1)(20 K)=2.16×10−4 m

Consider slab (e).

Substitute 60 cm for L, 50°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(12×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(12×10−6 K−1)(30 K)=2.16×10−4 m

Thus, the rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

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Answer 5

Chapter 13, Problem 6P

To determine

Rank the increase in length from greatest to smallest.

Expert Solution & Answer

Answer to Problem 6P

The rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Explanation of Solution

Write an expression for the increase in length.

ΔL=LαΔT=Lα(Tf−Ti) (I)

Here, ΔL is the increase in length, L is the initial length, α is the coefficient of thermal expansion, ΔT is the change in temperature, Tf is final temperature and Ti is the initial temperature.

Conclusion:

Consider slab (a).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(20 K)=1.44×10−4 m

Consider slab (b).

Substitute 90 cm for L, 50°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(30 K)=2.16×10−4 m

Consider slab (c).

Substitute 60 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(8×10−6 K−1)(20 K)=0.96×10−4 m

Consider slab (d).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(12×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(12×10−6 K−1)(20 K)=2.16×10−4 m

Consider slab (e).

Substitute 60 cm for L, 50°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(12×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(12×10−6 K−1)(30 K)=2.16×10−4 m

Thus, the rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Answer 6

Chapter 13, Problem 6P

To determine

Rank the increase in length from greatest to smallest.

Expert Solution & Answer

Answer to Problem 6P

The rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Explanation of Solution

Write an expression for the increase in length.

ΔL=LαΔT=Lα(Tf−Ti) (I)

Here, ΔL is the increase in length, L is the initial length, α is the coefficient of thermal expansion, ΔT is the change in temperature, Tf is final temperature and Ti is the initial temperature.

Conclusion:

Consider slab (a).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(20 K)=1.44×10−4 m

Consider slab (b).

Substitute 90 cm for L, 50°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(30 K)=2.16×10−4 m

Consider slab (c).

Substitute 60 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(8×10−6 K−1)(20 K)=0.96×10−4 m

Consider slab (d).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(12×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(12×10−6 K−1)(20 K)=2.16×10−4 m

Consider slab (e).

Substitute 60 cm for L, 50°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(12×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(12×10−6 K−1)(30 K)=2.16×10−4 m

Thus, the rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Answer 7

Chapter 13, Problem 6P

To determine

Rank the increase in length from greatest to smallest.

Answer 8

Expert Solution & Answer

Answer to Problem 6P

The rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write an expression for the increase in length.

ΔL=LαΔT=Lα(Tf−Ti) (I)

Here, ΔL is the increase in length, L is the initial length, α is the coefficient of thermal expansion, ΔT is the change in temperature, Tf is final temperature and Ti is the initial temperature.

Conclusion:

Consider slab (a).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(20 K)=1.44×10−4 m

Consider slab (b).

Substitute 90 cm for L, 50°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(8×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(8×10−6 K−1)(30 K)=2.16×10−4 m

Consider slab (c).

Substitute 60 cm for L, 40°C for Tf, 20°C for Ti and 8×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(8×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(8×10−6 K−1)(20 K)=0.96×10−4 m

Consider slab (d).

Substitute 90 cm for L, 40°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((90 cm)(1 m10−2 cm))(12×10−6 K−1)((40°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.90 m)(12×10−6 K−1)(20 K)=2.16×10−4 m

Consider slab (e).

Substitute 60 cm for L, 50°C for Tf, 20°C for Ti and 12×10−6 K−1 for α in equation (I) to find ΔL.

ΔL=((60 cm)(1 m10−2 cm))(12×10−6 K−1)((50°C+273.15 K/°C)−(20°C+273.15 K/°C))=(0.60 m)(12×10−6 K−1)(30 K)=2.16×10−4 m

Thus, the rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Answer 12

Ch. 13.2 - Practice Problem 13.1 Normal Body Temperatures...Ch. 13.3 - Prob. 13.3CP Ch. 13.3 - Prob. 13.2PP Ch. 13.3 - Prob. 13.3PP Ch. 13.4 - Prob. 13.4CP Ch. 13.4 - Prob. 13.4PP Ch. 13.5 - Prob. 13.5CP Ch. 13.5 - Prob. 13.5PP Ch. 13.5 - Prob. 13.6PP Ch. 13.6 - Prob. 13.6CP

Rank the increase in length from greatest to smallest.

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Rank the increase in length from greatest to smallest.

Answer to Problem 6P

Explanation of Solution

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